The constraints and their derivatives are formed by opf_consfcn.m.

-- 
Ray Zimmerman
Senior Research Associate
419A Warren Hall, Cornell University, Ithaca, NY 14853
phone: (607) 255-9645




On May 19, 2013, at 9:06 AM, Aftognosia Aftognosia <[email protected]> wrote:

> Dear Dr Zimmerman,
>  
> I would like to ask you, where is it possible to find the source code of the 
> matpower ACOPF equations?
> SInce you already mentioned that it is open to everyone!
>  
> Thank you
> 
> From: Ray Zimmerman <[email protected]>
> To: MATPOWER discussion forum <[email protected]> 
> Sent: Tuesday, May 14, 2013 7:32 PM
> Subject: Re: branch modeling
> 
> Yes.
> 
> On May 14, 2013, at 12:55 AM, Petre Razusi <[email protected]> wrote:
> 
>> Thank you Prof. Zimmerman. Your explanations helped me. However, I still 
>> have another thing that is a little bit unclear.
>> I have some transformer parameters (in Ohms and Siemes) calculated using the 
>> "to" side voltage. How do I convert them to p.u.? Do I use the same Zbase 
>> that you suggested?
>> 
>> Thank you for your help!
>> 
>> Petre
>> 
>> 
>> 2013/5/13 Ray Zimmerman <[email protected]>
>> Since the branch impedances are on the "to" side of the branch, you are 
>> correct that you would use the voltage base from the "to" bus.
>> 
>> The formula isn't really a MATPOWER question, so you should be able to find 
>> the answer in a good power systems textbook, however it is very 
>> straightforward to derive yourself. Essentially we want to find the 
>> impedance base (Zbase) in terms of the power and voltage bases (Sbase and 
>> Vbase).
>> 
>> Start with some definitions ...
>> 
>> (1)  V = Vpu * Vbase
>> (2)  S = Spu * Sbase
>> (3)  Z = Zpu * Zbase
>> 
>> Then some simple relationships ...
>> 
>> (4)  V = I * Z
>> (5)  S = V * conj(I)
>> 
>> Solving (4) for I and substituting in (5), then solving for Z, gives ...
>> 
>> (6) Z = |V|^2 / conj(S)
>> 
>> And we know the p.u. quantities have the same relationships …
>> 
>> (7) Zpu = |Vpu|^2 / conj(Spu)
>> 
>> Substituting (1)-(3) and then (7) into (6) and rearranging, gives …
>> 
>> Zbase = Vbase^2 / Sable
>> 
>> Hope this helps,
>> 
>> --
>> Ray Zimmerman
>> Senior Research Associate
>> 419A Warren Hall, Cornell University, Ithaca, NY 14853
>> phone: (607) 255-9645
>> 
>> 
>> 
>> 
>> On May 13, 2013, at 3:44 AM, Petre Razusi <[email protected]> wrote:
>> 
>> > Dear all,
>> >
>> > I have some things I don't quite understand regarding the modeling of 
>> > branches and the pu system.
>> > If I'm not wrong, to transform the R,X and B parameters to pu one needs a 
>> > base impedance. Since the model used for the branch is a complex one, 
>> > dealing with both transformers and lines, I presume that this impedance is 
>> > calculated using the "to" bus voltage. Am I right? And if so, could you 
>> > tell me what is the formula used to compute the base impedance?
>> >
>> > Thanks,
>> >
>> > Petre
>> 
>> 
>> 
>> 
>> 
>> 
>> 
> 
> 
> 

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